Problem 73 - Calculus for Biology and Medicine

Question

73\. An urn contains 1 black and 9 white balls. Balls are drawn at random until the black ball is selected. Find the probability that exactly 6 white balls will be drawn before the black one is if (a) each ball is replaced before the next ball is drawn and (b) balls are not replaced.

Step-by-Step Solution

Answer

(a) \( \frac{9^6}{10^7} \); (b) \( \frac{1}{10} \).

Step 1: Understand the Problem

We have an urn with 1 black and 9 white balls. We need to determine the probability that exactly 6 white balls are drawn before the black one is selected. We will evaluate this with replacement and without replacement.

Step 2: Calculate With Replacement Probability

When balls are replaced: Each draw is an independent event with a probability of drawing a white ball as \( \frac{9}{10} \) and a black ball as \( \frac{1}{10} \). For exactly 6 white balls to precede the black ball, you must draw 6 white balls and then the black ball on the 7th draw. The probability for this sequence is \( \left( \frac{9}{10} \right)^6 \times \frac{1}{10} \), which evaluates to \( \frac{9^6}{10^7} \).

Step 3: Calculate Without Replacement Probability

When balls are not replaced: Initially, there are 10 balls. The probability sequence must involve drawing 6 white balls first and then 1 black ball from the remaining 4 balls (3 white and 1 black). For this, the probability is calculated by multiplying probabilities at each draw: \( \left( \frac{9}{10} \right) \times \left( \frac{8}{9} \right) \times \left( \frac{7}{8} \right) \times \left( \frac{6}{7} \right) \times \left( \frac{5}{6} \right) \times \left( \frac{4}{5} \right) \times \left( \frac{1}{4} \right) \), which simplifies to \( \frac{1}{10} \).

Key Concepts

Understanding Replacement Events

Replacement events are an essential concept in probability theory. It refers to the scenario where each item drawn from a set is returned to the set before the next item is drawn. This means the composition of the set remains constant for each draw.
In the given exercise, replacement affects the probability of drawing a white or black ball. Since the balls are replaced, each draw is independent. The probability of drawing a white ball on each attempt is always \( \frac{9}{10} \), while drawing a black ball remains \( \frac{1}{10} \).
This consistency in probability streamlines calculations and simplifies the process of finding exact sequences of events, such as drawing a specified number of white balls before a black one.

Concept of Independent Events

Independent events are those where the outcome of one event does not affect the outcome of another. In the context of replacement events, each draw of a ball is independent as the replacement keeps probabilities unchanged.
This property simplifies probability calculations, allowing the probabilities of several independent events to be multiplied together to find combined probabilities.
In the exercise, since each draw is independent when events involve replacement, we calculate the probability for 6 white balls and a black ball using the formula:

• \( \left( \frac{9}{10} \right)^6 \times \frac{1}{10} \)

This multiplication illustrates how independent events are dealt with in probability theory.

Exploring Conditional Probability

Conditional probability considers the probability of an event given that another event has already occurred. This comes into play when not replacing the drawn ball, where the outcome of one draw affects the subsequent draws.
When a white ball is drawn, that alters the composition of balls for the next draw, changing the probability of drawing a white or black ball next. The exercise involves calculating probabilities under changing conditions.
Initially, for drawing 6 white balls followed by a black one, each draw's probability changes as balls are not replaced:

• Start with \( \frac{9}{10} \)
• Then \( \frac{8}{9} \), and so forth
• Until hitting \( \frac{1}{4} \) for the last black ball

This showcases conditional probability's role in adjusting calculations based on changing conditions.

Basics of Combinatorics

Combinatorics deals with counting, arranging, and combination possibilities. It plays an integral role in probability, especially when sequences or arrangements matter.
In the exercise, combinatorics helps determine the sequence: drawing exactly 6 white balls before one black ball.
It involves considering different potential outcomes and permutations, such as:

• All combinations where 6 white balls appear before the black one
• By employing probability formulas efficiently to streamline complex combinations

Understanding these basics makes it easier to untangle seemingly complicated probability problems.